The difference between BigN and the n of specific records is going to have one prime number as its largest prime factor. e.g. c=501 bign=229 n=63 gap=166 LPF=83 (gap=83*2) {17:63:22:19:3:167}
Multiple records in the grid have different gaps with the same highest prime factor, e.g. c=835 bign=390 n=58 gap=332 LPF=83 (8322) (51,58,12) = {51:58:28:23:5:167}
Since BigN and n are always the same parity, this gap will always be even.
Since multiple records have different gaps with the same highest prime factor, they can be grouped together, like in pic related.
The beginning record always has a=3 but b doesn’t seem predictable that I can tell.
The difference in the gap from one record to the next within a group is equal to the biggest prime factor times 2.
c=357 gap=118 BPF=59
c=595 gap=236 BPF=59 (236-118=118)
c=833 gap=354 BPF=59 (354-236=118)
c=1071 gap=472 BPF=59 (472-354=118)
c=1309 gap=590 BPF=59 (590-472=118)
This would only help if we could predict the BPF group any number falls within, because then we’d be able to use the way in which c scales upwards to calculate how many times we’d need to multiply the BPF by to get the correct BigN-n gap. Then we’d just take that gap away from the BigN we know how to calculate, and that would be it. Maybe there’s some way of finding that BPF.